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J OURNAL DE T HÉORIE DES N OMBRES DE B ORDEAUX TAKAO KOMATSU The fractional part of nθ + and Beatty sequences Journal de Théorie des Nombres de Bordeaux, tome p. 387-406 7, no 2 (1995), <http://www.numdam.org/item?id=JTNB_1995__7_2_387_0> © Université Bordeaux 1, 1995, tous droits réservés. L’accès aux archives de la revue « Journal de Théorie des Nombres de Bordeaux » (http://jtnb.cedram.org/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/legal.php). Toute utilisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright. Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques http://www.numdam.org/ 387 The fractional part of n03B8 + ~ and Beatty sequences par TAKAO KOMATSU 1. Introduction be the denominator of the n-th convergent of the Let () be real and for continued fraction expansion of 0. Thus when 0 is rational, 8 some integer N. Denote the fractional part of 0 by {9}, the floor, that is and its distance from the nearest integer by ~~8~~. integer part, of 0 by = It is well known that the continued fraction convergents imations to 0 in the sense that are best approx- corresponding results for one-sided best approximations are readily derived, see for example van Ravenstein [9j, and are ’ The ° and The purpose of this paper is to give similar results in the inhomogeneous x8 -- , where 0 and ø are real numbers, not necessarily irrational, and to apply these results to find the characteristic word case of the Beatty sequence. A Beatty sequence is a sequence of the form for fixed real numbers 0 and 0. The homogeneous case § = 0 has been dealt 1991 Mathematics Subject Cla,ssi. fic.a~tion. 11B83.. Mots clés : continued fraction, Beatty sequence. Manuscrit re~u le 23 mai 1994 388 with extensively (see for example [3], [4], [5], [8], [10], [11], [12], and [2] 0 is also discussed for additional references). The inhomogeneous from several viewpoints (see [5], [7J, [8]), but these sources do not necessarily make it easy to get the actual Beatty sequences. The paper [5] is complete on the theory, but only the limited periodic case yields Beatty sequences. The paper [8] provides a very effective description of Beatty sequences, but regrettably, it is now known that the results do not match the facts (see the corrected version, (6~). 2. The minimum of the fractional We first consider how to find an part of z0 + § integer x satisfying When both 0 and 0 are rational, the problem is not too difficult. But how should we treat the irrational case? The answer is given in the following theorem. Using this result, we can find x directly in every case - though two values must be compared when 8 is irrational. be real THEOREM 1. 0. Then n for >_ 2, vergent of Proof. Set an When n is odd, = numbers, and let Consider the pn/qn. .denote the n-th We recall that con- 389 To minimize {xB + 01, And this is possible since Now, need we (pn, qn) is since a = 1 and o x qn. solution of is a solution of the diophantine equa- tion Thus, the solutions of this equation are we get Therefore, when n is odd and In like manner, when Similarly, Thus, n (x pn + an)/qn is odd and when n is even, E Z, (xpn + an - 1)/qn E Z, 390 that is When 0 = pn /qn for some n > 2, Now gives Rerriark. If a particular disallow x 0 (in order that van Ravenstein’s result case), Theorem 1 must be modified. In this case we = [9] is ’ when 0 Corresponding results for the maximum can be obtained symmetrically: One uses the ceiling instead of the floor in the proof above. COROLLARY. Let 8, ~ be any real numbers and let vergent of 0. Then for n > 2 the n-th and t=o. Remark. In the range 0 x qn in which x = 0 is disallowed, con- 391 3. THE MINIMUM VALUE OF + oil Combining the results for minimum and maximum, we get the following theorem: be any Teal numbers and let pn/qn be the n-th THEOREM 2. 0. Then vergent of for n > 2 where ~~, +t) _ 1 or 2 When 0 = if n is mod qn and t = -1, 0 or 1 if n is con- odd; even. (n > 2), llxo ~- Oil has period qn in x and andt=0 orl. 0 or we have t Furthermore, if an.~l _> 2, then for any real 0 and for infinitely many n with r~ > 2 we have And for any real 0 = 1. t=0 orl. Proof. The first two assertions are obvious. when n It suffices to deal with the case are three cases, say is odd. Then from the first assertion there We will show that Since and 392 it follows that On the other hand if this value is positive, otherwise. Since we have From (1) and conclusion. (2) the fact qn+l = 2q~ + now yields the , Using Hurwitz’s theorem for infinitely pn/qn, many that is for infinitely many n and we obtain the last part of Theorem 2. qn, in which Remark.. In the range 0 x 4. The third in Theorem 2 possibility x = 0 is disallowed, If 0 is irrational and an+l = 1 for some n~> 2), there may be the third possibility, namely t -1 when n is odd; t 2 when n is even. In this section we describe () and § which give this third possibility. = = 393 Once again it suffices to consider just odd n. Again set an - Lqnoj -_ and + t) mod qn, where t -1, 0, 1. Then, the third if occurs possibility = But (3) is equivalent to or where x where c an+l = = ~~-1). tively for odd n in By comparing with we must satisfy c 2, which we determine later. Let positive integers, say Ln and L~.~2, respecthe continued fraction expansion of 0. Then where Since 0 x qn - 1, 8 a constant with 1 an and an+2 be large is 1, 0 and Since 0 - may take an+2, - - -] (5), we get Ln+2 This condition will be as and 0 large as E 1. Thus, this 0 satisfies like, the condition we on c ~5) . is sharpened in (8). We still have to make x = 1) mod qn large enough and 0 small enough to satisfy (4). First, we select x, have to make § To satisfy the conditions and then may choose knpn -E-1, z of (4), the value of must be in the range = = 394 Together with the condition for This condition will be Next, we select 0. We sharpened can equivalent whence To see that, whence we we have note to in get (7). say that the where the two fractions are since qn > qn the definition of an, we get This is c we integers. We conclude that condition (3) = x is as large as is qn - kn and given the conditions we on have kn. Recalling 395 and Among these kn satisfying (6) we choose kn such that the left hand side of inequality defining the range of 0 is less than the right hand side, so the that is, When n > 3, it follows from We remark that 2(qn near to 2 because L~, is When there is we can select 0 a (5) that such kn exist if we take 1)/(2qn - large. kn satisfying (7) is near to 1 and c as 2(qn - is , and as The 0 and 0 so selected give the third even, instead of (9) we can choose 0 as possibility for odd By these steps, n. When n is we can always find 0 and § to give the third possibility for odd or 1, by taking Ln and Ln+2 large n, any any even n when an+, so enough that there is an integer kn satisfying the conditions (7) and (9) (or (9’)). Moreover, it is possible to obtain § giving the third possibility for infinitely many odd (or even) n’s. = 396 show the way to find the next interval such that In+2 C In: with 0 6n Take an increasing sequence of constants 1/3 for all we shall odd n choose For each n. (> 3) always positive integers We now and where L is an integer with L > 8 and kn satisfying condition (7). Put and where and using the we assume notation of the ceiling that and Then, comparing with (7), and ~3n 1 -1/2qn yields (6). Let In be the interval of (10). Note that if will automatically satisfy condition (11). In+2 C I~, then We examine the fractional parts of various integer multiples of Pn+2/ qn+2 with an eye to combining them to find an integer kn+2 such that is in so a subinterval of First, 397 We also have and where the third equality (12) and (14) we have Now let 1 be the Using (13) and we can be obtained by adding the first non-negative integer satisfying get ’ two. From 398 for some 1 with where i Now Having chosen and,8,, in (10). Therefore, and In view define an+2 and and (16) we have we of (15) #n+2 analogously to an C In - give the third possibility for each odd EXAMPLES. give the third possibility for r give the third possibility for n 5. The sorting of = 2, 4, 6, 8, - - - . . {x8 + 0} We can solve the sorting problem by using the former results. When 0 is irrational, let qn be the denominator of the n-th convergent of 0, as above. When 0 is rational, let 0 p~/qn, where pn and qn (> 0) are + ~~), j - 1, 2, - - - , N, be the ordered sequence of coprime. Let fractional parts. That is, Jul, uZ, - - - ,) 10, 1, 2, - - - , N - 1 ) , and = = When 0 is irrational and ¢ = 0, the following result is well known: 399 When N LEMMA. where u2 Proof. = qn, for j if n is odd, = This follows - u2 = 1, 2, ..., if n is qn - immediately from Lemma 2.1 THEOREM 3. When N where t (-1)n 0 or = qn , and is even. and Theorem 3.3 of for j - 1, 2, ... , qn independent of j. When x8 + 0 is not integer, for each Proof. Set an integer t satisfying Let x(t) = x~t~ with 0 x(t) qn satisfy Then From Theorem 1 t must be 0 or (-1)". When x6 + 0 is integer, there exists In particular, t = [9]. which includes the result above. We will sort the fractional part = qn 0 when 0 is rational, or satisfying if o ~ 1/qn . x there exists 400 to the Application 6. inhomogeneous Beatty sequence we introduce the easiest method for getting the inhoBeatty sequences in every case. Indeed, one easily obtain the sequence by the use our theorems once one knows the corresponding homogeneous Beatty sequences; see the examples in §7. Consider the sequence of differences, that is the characteristic sequence In this section, mogeneous First, let 0 be rational. Then, the following theorem Let 0 be THEOREM 4. an a rational number, 0 real a is essential. number, and let r~ be integer satisfying Then for all integer n we have Proof. Plainly Because ( we have Since also the result follows. Next, we consider the case () irrational. The main theorem is Theorem which will follow from Proposition 1 and Proposition 2. The method of 5, proof is similar to that of the rational case. As seen in the assertions, we need not necessarily take an integer x which minimizes the fractional part of z0 + 0, but it is surely the best way to use such an integer. PROPOSITION 1. be an irrational number with continued fraction expansion - and let 0 by pilqi . If n with be a real number. For i > - = 0,1, 2, ... denote the convergents of 0 f ~o+~~ for a non-negative integer k, then for all integer qi - 2 we have f (n -~- J~; 8, ~) = f ~n; e, o~ . This proof depends on the following lemmata: 401 LEMMA 1. Let 0 be + tive odd ~} integer i and for every positive Proof. irrational number and 0 any real number. If for a non-negative integer k, then for every posi- an When i is even integer i odd, Then, Since together with the first inequality follows. When i is even, and Then, 402 Since together with the third inequality is proved. The second and the fourth LEMMA 2. If every integer n with inequalities are proved similarly. {~8 + 01 for a non-negative integers k, Inj then for qi , Proof. First let n > o. By the result of [9] and the first and the third inequalities in Lemma 1, Lemma 2 holds for n which maximizes the fractional part when 0 n qi for integers i > 2, whence together with the the lemma holds for the other positive n. When n - 0, the result is trivial. Finally, if n 0, by [9] and the second and the fourth inequalities in Lemma 1, for 0 n’ qi, condition (x0 + 01 Hence, together with we have the claim. Proof of Proposition 1 ~ If {kB + ol By Lemma 2, InO + (kB + 0)} the number line to the left of integer n with Ini qi = 0, then for any integer = {nB + {kB + ~}} is never n located on the Therefore, if (k0 + ~} ~ 0, for each 403 This gives As the possibility inequality of Lemma 2 may not hold for n qa, there is the = that be PROPOSITION 2. an irrational number with continued fraction expansion and let 0 be a real number. Denote by pilqi (i vergent of the continued fraction expansion of 0. integer satisfying If l > i for an integer i with i > = 1,2,...) Let r~l be 2, then for all integers a n the i-th con- non-negative with Proof. This is similar to the proof of Theorem 4. Now, we combine Propositions 1 and 2 to get the main theorem for an irrational 0. THEOREM 5. Let 0 be acn irrational number wzth continued fraction expan- sion real number. Denote by pilqi (i 1,2, ...) the i-th convergent the continued fraction expansion of 0. Let "’I be a non-negative integer a = of satisfying then for all integers n with 404 Remark. We don’t have to care for the mysterious condition For, always in these 7. cases. Examples and by Theorem 1 From and we we get ~4 20. Next, since by Theorem 5 = have For example, according to the method of Fraenkel and we get et. al. [4], as 405 Furthermore, and that is is where the underline indicates the place f (0; V2, 0). and by Theorem 1 Next, by Theorem 4 we have for all integer For example, according to the method of Fraenkel et. al. n . [4~, as and . we get Therefore, f (n; 36/25, for n E Z is where the underline indicates the place f indicates the place f (0; 36/25, Of course, we can use the form characteristic sequence. 8. (0; 36/25, 0) and the double-underline 36/25 = (1, 2, 3,1,1) to have the same Acknowledgment I wish to thank Dr P. A. B. Pleasants and Professor A. J. van der Poorten for their advice and assistance in my preparation of the present paper. 406 REFERENCES [1] J. M. BORWEIN and P. B. BORWEIN, On the generating function part: [n03B1 + 03B3], J. Number Theory 43 (1993), 293-318. [2] T. C. BROWN, Descriptions of the characteristic sequence Math. Bull 36 (1993), 15-21. [3] L. V. DANILOV, Some class of transcendental 149-154= Math. Notes 12 (1972), 524-527. of the integer of an irrational, Canad. numbers, Mat. Zametki 12 (1972), FRAENKEL, M. MUSHKIN and U. TASSA, Determination of [n03B8] by its of differences, Canad. Math. Bull. 21 (1978), 441-446. Sh. ITO and S. YASUTOMI, On continued fractions, substitutions and character[5] istic sequences [nx + y] - [(n - 1)x + y], Japan. J. Math. 16 (1990), 287-306. [6] T. KOMATSU, A certain power series associated with Beatty sequences, manuscript. [7] T. KOMATSU, On the characteristic word of the inhomogeneous Beatty sequence, [4] A. S. sequence Bull. Austral. Math. Soc. 51 [8] K. (1995), 337-351. NISHIOKA, I. SHIOKAWA and J. TAMURA, Arithmetical properties of certain series, J. Number Theory 42 (1992), 61-87. VAN RAVENSTEIN, The three gap theorem (Steinhaus conjecture), J. Austral. power [9] T. Math. Soc. (Series A) 45 (1988), 360-370. [10] T. VAN RAVENSTEIN, G. WINLEY and K. TOGNETTI, Characteristics and the three gap theorem, Fibonacci Quarterly 28 (1990), 204-214. [11] K. B. STOLARSKY, Beatty sequences, continued fractions, and certain ators, Canad. Math. Bull. 19 (1976) , 473-482. [12] B. A. shift oper- VENKOV, Elementary Number Theory, Wolters-Noordhoff, Groningen, 1970. Takao KOMATSU School of Mathematics, Physics, Computing and Electronics Macquarie University NSW 2109 Australia e--mail : [email protected]